3.78.48 \(\int \frac {-512+500 x-78 x^2-4 x^3+e^2 (4 x^3-x^4) \log (4)+(32-32 x+6 x^2) \log (4-x)}{e^2 (-16 x^3+20 x^4-8 x^5+x^6)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\log (4)+\frac {2 (16+x-\log (4-x))}{e^2 x^2}}{-2+x} \]

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Rubi [B]  time = 0.85, antiderivative size = 89, normalized size of antiderivative = 3.18, number of steps used = 25, number of rules used = 11, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {12, 6741, 6742, 44, 88, 72, 2418, 2395, 36, 31, 29} \begin {gather*} -\frac {16}{e^2 x^2}+\frac {\log (4-x)}{e^2 x^2}-\frac {9}{e^2 (2-x)}-\frac {9}{e^2 x}+\frac {\log (4-x)}{2 e^2 (2-x)}+\frac {\log (4-x)}{2 e^2 x}-\frac {\log (4)}{2-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-512 + 500*x - 78*x^2 - 4*x^3 + E^2*(4*x^3 - x^4)*Log[4] + (32 - 32*x + 6*x^2)*Log[4 - x])/(E^2*(-16*x^3
+ 20*x^4 - 8*x^5 + x^6)),x]

[Out]

-9/(E^2*(2 - x)) - 16/(E^2*x^2) - 9/(E^2*x) - Log[4]/(2 - x) + Log[4 - x]/(2*E^2*(2 - x)) + Log[4 - x]/(E^2*x^
2) + Log[4 - x]/(2*E^2*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-512+500 x-78 x^2-4 x^3+e^2 \left (4 x^3-x^4\right ) \log (4)+\left (32-32 x+6 x^2\right ) \log (4-x)}{-16 x^3+20 x^4-8 x^5+x^6} \, dx}{e^2}\\ &=\frac {\int \frac {512-500 x+78 x^2+4 x^3-e^2 \left (4 x^3-x^4\right ) \log (4)-\left (32-32 x+6 x^2\right ) \log (4-x)}{(2-x)^2 (4-x) x^3} \, dx}{e^2}\\ &=\frac {\int \left (-\frac {4}{(-4+x) (-2+x)^2}-\frac {512}{(-4+x) (-2+x)^2 x^3}+\frac {500}{(-4+x) (-2+x)^2 x^2}-\frac {78}{(-4+x) (-2+x)^2 x}-\frac {e^2 \log (4)}{(-2+x)^2}+\frac {2 (-4+3 x) \log (4-x)}{(-2+x)^2 x^3}\right ) \, dx}{e^2}\\ &=-\frac {\log (4)}{2-x}+\frac {2 \int \frac {(-4+3 x) \log (4-x)}{(-2+x)^2 x^3} \, dx}{e^2}-\frac {4 \int \frac {1}{(-4+x) (-2+x)^2} \, dx}{e^2}-\frac {78 \int \frac {1}{(-4+x) (-2+x)^2 x} \, dx}{e^2}+\frac {500 \int \frac {1}{(-4+x) (-2+x)^2 x^2} \, dx}{e^2}-\frac {512 \int \frac {1}{(-4+x) (-2+x)^2 x^3} \, dx}{e^2}\\ &=-\frac {\log (4)}{2-x}+\frac {2 \int \left (\frac {\log (4-x)}{4 (-2+x)^2}-\frac {\log (4-x)}{x^3}-\frac {\log (4-x)}{4 x^2}\right ) \, dx}{e^2}-\frac {4 \int \left (\frac {1}{4 (-4+x)}-\frac {1}{2 (-2+x)^2}-\frac {1}{4 (-2+x)}\right ) \, dx}{e^2}-\frac {78 \int \left (\frac {1}{16 (-4+x)}-\frac {1}{4 (-2+x)^2}-\frac {1}{16 x}\right ) \, dx}{e^2}+\frac {500 \int \left (\frac {1}{64 (-4+x)}-\frac {1}{8 (-2+x)^2}+\frac {1}{16 (-2+x)}-\frac {1}{16 x^2}-\frac {5}{64 x}\right ) \, dx}{e^2}-\frac {512 \int \left (\frac {1}{256 (-4+x)}-\frac {1}{16 (-2+x)^2}+\frac {1}{16 (-2+x)}-\frac {1}{16 x^3}-\frac {5}{64 x^2}-\frac {17}{256 x}\right ) \, dx}{e^2}\\ &=-\frac {9}{e^2 (2-x)}-\frac {16}{e^2 x^2}-\frac {35}{4 e^2 x}-\frac {\log (4)}{2-x}+\frac {\log (2-x)}{4 e^2}-\frac {\log (4-x)}{16 e^2}-\frac {3 \log (x)}{16 e^2}+\frac {\int \frac {\log (4-x)}{(-2+x)^2} \, dx}{2 e^2}-\frac {\int \frac {\log (4-x)}{x^2} \, dx}{2 e^2}-\frac {2 \int \frac {\log (4-x)}{x^3} \, dx}{e^2}\\ &=-\frac {9}{e^2 (2-x)}-\frac {16}{e^2 x^2}-\frac {35}{4 e^2 x}-\frac {\log (4)}{2-x}+\frac {\log (2-x)}{4 e^2}-\frac {\log (4-x)}{16 e^2}+\frac {\log (4-x)}{2 e^2 (2-x)}+\frac {\log (4-x)}{e^2 x^2}+\frac {\log (4-x)}{2 e^2 x}-\frac {3 \log (x)}{16 e^2}-\frac {\int \frac {1}{(4-x) (-2+x)} \, dx}{2 e^2}+\frac {\int \frac {1}{(4-x) x} \, dx}{2 e^2}+\frac {\int \frac {1}{(4-x) x^2} \, dx}{e^2}\\ &=-\frac {9}{e^2 (2-x)}-\frac {16}{e^2 x^2}-\frac {35}{4 e^2 x}-\frac {\log (4)}{2-x}+\frac {\log (2-x)}{4 e^2}-\frac {\log (4-x)}{16 e^2}+\frac {\log (4-x)}{2 e^2 (2-x)}+\frac {\log (4-x)}{e^2 x^2}+\frac {\log (4-x)}{2 e^2 x}-\frac {3 \log (x)}{16 e^2}+\frac {\int \frac {1}{4-x} \, dx}{8 e^2}+\frac {\int \frac {1}{x} \, dx}{8 e^2}-\frac {\int \frac {1}{4-x} \, dx}{4 e^2}-\frac {\int \frac {1}{-2+x} \, dx}{4 e^2}+\frac {\int \left (-\frac {1}{16 (-4+x)}+\frac {1}{4 x^2}+\frac {1}{16 x}\right ) \, dx}{e^2}\\ &=-\frac {9}{e^2 (2-x)}-\frac {16}{e^2 x^2}-\frac {9}{e^2 x}-\frac {\log (4)}{2-x}+\frac {\log (4-x)}{2 e^2 (2-x)}+\frac {\log (4-x)}{e^2 x^2}+\frac {\log (4-x)}{2 e^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.16, size = 87, normalized size = 3.11 \begin {gather*} \frac {128+8 x+2 (-2+x) x^2 \tanh ^{-1}(3-x)+4 e^2 x^2 \log (4)+(-2+x) x^2 \log (2-x)-8 \log (4-x)+2 x^2 \log (4-x)-x^3 \log (4-x)}{4 e^2 (-2+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-512 + 500*x - 78*x^2 - 4*x^3 + E^2*(4*x^3 - x^4)*Log[4] + (32 - 32*x + 6*x^2)*Log[4 - x])/(E^2*(-1
6*x^3 + 20*x^4 - 8*x^5 + x^6)),x]

[Out]

(128 + 8*x + 2*(-2 + x)*x^2*ArcTanh[3 - x] + 4*E^2*x^2*Log[4] + (-2 + x)*x^2*Log[2 - x] - 8*Log[4 - x] + 2*x^2
*Log[4 - x] - x^3*Log[4 - x])/(4*E^2*(-2 + x)*x^2)

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fricas [A]  time = 0.54, size = 34, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left (x^{2} e^{2} \log \relax (2) + x - \log \left (-x + 4\right ) + 16\right )} e^{\left (-2\right )}}{x^{3} - 2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2-32*x+32)*log(-x+4)+2*(-x^4+4*x^3)*exp(2)*log(2)-4*x^3-78*x^2+500*x-512)/(x^6-8*x^5+20*x^4-16
*x^3)/exp(2),x, algorithm="fricas")

[Out]

2*(x^2*e^2*log(2) + x - log(-x + 4) + 16)*e^(-2)/(x^3 - 2*x^2)

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giac [A]  time = 0.19, size = 34, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left (x^{2} e^{2} \log \relax (2) + x - \log \left (-x + 4\right ) + 16\right )} e^{\left (-2\right )}}{x^{3} - 2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2-32*x+32)*log(-x+4)+2*(-x^4+4*x^3)*exp(2)*log(2)-4*x^3-78*x^2+500*x-512)/(x^6-8*x^5+20*x^4-16
*x^3)/exp(2),x, algorithm="giac")

[Out]

2*(x^2*e^2*log(2) + x - log(-x + 4) + 16)*e^(-2)/(x^3 - 2*x^2)

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maple [A]  time = 0.13, size = 43, normalized size = 1.54

method	result	size
norman	\(\frac {2 x^{2} \ln \relax (2)+32 \,{\mathrm e}^{-2}+2 x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2} \ln \left (-x +4\right )}{\left (x -2\right ) x^{2}}\)	\(43\)
risch	\(-\frac {2 \,{\mathrm e}^{-2} \ln \left (-x +4\right )}{x^{2} \left (x -2\right )}+\frac {2 \,{\mathrm e}^{-2} \left (x^{2} {\mathrm e}^{2} \ln \relax (2)+x +16\right )}{\left (x -2\right ) x^{2}}\)	\(43\)
default	\({\mathrm e}^{-2} \left (-\frac {2 \,{\mathrm e}^{2} \ln \relax (2)}{2-x}-\frac {9}{x}-\frac {\ln \left (-x +4\right ) \left (-x +4\right ) \left (-x -4\right )}{16 x^{2}}+\frac {\ln \left (-x +4\right ) \left (-x +4\right )}{8 x}+\frac {\left (-x +4\right ) \ln \left (-x +4\right )}{-4 x +8}-\frac {9}{2-x}-\frac {\ln \left (-x +4\right )}{16}-\frac {16}{x^{2}}\right )\)	\(104\)
derivativedivides	\(-{\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e}^{2} \ln \relax (2)}{2-x}+\frac {9}{x}+\frac {\ln \left (-x +4\right ) \left (-x +4\right ) \left (-x -4\right )}{16 x^{2}}-\frac {\ln \left (-x +4\right ) \left (-x +4\right )}{8 x}-\frac {\ln \left (-x +4\right ) \left (-x +4\right )}{4 \left (2-x \right )}+\frac {9}{2-x}+\frac {\ln \left (-x +4\right )}{16}+\frac {16}{x^{2}}\right )\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2-32*x+32)*ln(-x+4)+2*(-x^4+4*x^3)*exp(2)*ln(2)-4*x^3-78*x^2+500*x-512)/(x^6-8*x^5+20*x^4-16*x^3)/ex
p(2),x,method=_RETURNVERBOSE)

[Out]

(2*x^2*ln(2)+32/exp(2)+2*x/exp(2)-2/exp(2)*ln(-x+4))/(x-2)/x^2

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maxima [B]  time = 0.51, size = 78, normalized size = 2.79 \begin {gather*} 2 \, {\left (\frac {{\left (e^{2} \log \relax (2) + 36\right )} x^{2} + {\left (x^{3} - 2 \, x^{2} - 1\right )} \log \left (-x + 4\right ) - 31 \, x}{x^{3} - 2 \, x^{2}} - \frac {4 \, {\left (9 \, x^{2} - 8 \, x - 4\right )}}{x^{3} - 2 \, x^{2}} - \log \left (x - 4\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2-32*x+32)*log(-x+4)+2*(-x^4+4*x^3)*exp(2)*log(2)-4*x^3-78*x^2+500*x-512)/(x^6-8*x^5+20*x^4-16
*x^3)/exp(2),x, algorithm="maxima")

[Out]

2*(((e^2*log(2) + 36)*x^2 + (x^3 - 2*x^2 - 1)*log(-x + 4) - 31*x)/(x^3 - 2*x^2) - 4*(9*x^2 - 8*x - 4)/(x^3 - 2
*x^2) - log(x - 4))*e^(-2)

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mupad [B]  time = 0.42, size = 49, normalized size = 1.75 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{-2}\,\left (4\,\ln \left (4-x\right )-64\right )}{2}-2\,x\,{\mathrm {e}}^{-2}+x^2\,\left ({\mathrm {e}}^{-2}-{\mathrm {e}}^{-2}\,\left (2\,{\mathrm {e}}^2\,\ln \relax (2)+1\right )\right )}{x^2\,\left (x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(500*x + log(4 - x)*(6*x^2 - 32*x + 32) - 78*x^2 - 4*x^3 + 2*exp(2)*log(2)*(4*x^3 - x^4) - 512))
/(16*x^3 - 20*x^4 + 8*x^5 - x^6),x)

[Out]

-((exp(-2)*(4*log(4 - x) - 64))/2 - 2*x*exp(-2) + x^2*(exp(-2) - exp(-2)*(2*exp(2)*log(2) + 1)))/(x^2*(x - 2))

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sympy [B]  time = 0.81, size = 56, normalized size = 2.00 \begin {gather*} - \frac {- 2 x^{2} e^{2} \log {\relax (2 )} - 2 x - 32}{x^{3} e^{2} - 2 x^{2} e^{2}} - \frac {2 \log {\left (4 - x \right )}}{x^{3} e^{2} - 2 x^{2} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2-32*x+32)*ln(-x+4)+2*(-x**4+4*x**3)*exp(2)*ln(2)-4*x**3-78*x**2+500*x-512)/(x**6-8*x**5+20*x
**4-16*x**3)/exp(2),x)

[Out]

-(-2*x**2*exp(2)*log(2) - 2*x - 32)/(x**3*exp(2) - 2*x**2*exp(2)) - 2*log(4 - x)/(x**3*exp(2) - 2*x**2*exp(2))

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